Question

[Thuhuyen Le]

Bài1Tìm giá trị lớn nhất:

a)x-x^2

b)5-8x-x^2

d)5-x^2+2x-4y^2-4y

Bài 2 Tìm giá trị nhỏ nhất

x^2-4xy+5y^2+10x-22y+28

Answer 1

[OoO Pipy OoO]

\(A=x-x^2=-\left(x^2-2\times x\times\frac{1}{2}+\left(\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^2\right)=-\left[\left(x-\frac{1}{2}\right)^2-\frac{1}{4}\right]\)

\(\left(x-\frac{1}{2}\right)^2\ge0\)

\(\left(x-\frac{1}{2}\right)^2-\frac{1}{4}\ge-\frac{1}{4}\)

\(-\left[\left(x-\frac{1}{2}\right)^2-\frac{1}{4}\right]\le\frac{1}{4}\)

Vậy Max A = \(\frac{1}{4}\) khi x = \(\frac{1}{2}\)

***

\(B=5-8x-x^2=-\left(x^2+2\times x\times4+4^2-4^2-5\right)=-\left[\left(x+4\right)^2-21\right]\)

\(\left(x+4\right)^2\ge0\)

\(\left(x+4\right)^2-21\ge-21\)

\(-\left[\left(x+4\right)^2-21\right]\le21\)

Vậy Max B = 21 khi x = - 4 

***

\(C=5-x^2+2x-4y^2-4y=-\left(x^2-2\times x\times1+1^2-1^2+\left(2y\right)^2-2\times2y\times1+1^2-1^2-5\right)=-\left[\left(x-1\right)^2+\left(2y-1\right)^2-7\right]\)

\(\left(x-1\right)^2\ge0\)

\(\left(2y-1\right)^2\ge0\)

\(\left(x-1\right)^2+\left(2y-1\right)^2-7\ge-7\)

\(-\left[\left(x-1\right)^2+\left(2y-1\right)^2-7\right]\le7\)

Vậy Max C = 7 khi x = 1 và y = \(\frac{1}{2}\)